In geometry, Cylinder is a surface traced by a straight lines through a given curve and being parallel to a fixed direction. In this definition of cylinder,
- straight line is called generator
- guiding curve is called directrix
- fixed direction is called axis
Cylindrical helix (or helix) is a curve traced out on a surface of cylinder and cuts the generators at constant angle. The axis of cylinder is also known as axis of the helix.
If \( \vec{a}\)is unit vector along direction of generator and \( \vec{t}\) is
unit vector along tangent to the helix, then
\( \vec{t} \) and \( \vec{a}\) makes constant angle (say \( \alpha \)),
i.e., \( \vec{t}.\vec{a}=cos\alpha \)
Necessary and sufficient condition for a curve to be a helix is that its curvature and torsion are in constant ratio.
Proof
Let \( C:\vec{r}=\vec{r}( s )\) is helix in which \( \vec{a}\) is unit vector along direction of generator and \( \vec{t}\) is unit vector along tangent to the helix, then
the necessary and sufficient condition for a curve to be a helix is that
\( \vec{t}.\vec{a}=cos\alpha \) (i)
Differentiating (i) w. r. to.\( s\), we get
\( \kappa \vec{n}.\vec{a}=0\)
⟺\( \vec{n}.\vec{a}=0\) [In general\( \kappa \ne 0\)] (ii)
Again, differentiating (ii) w. r. to.\( s\), we get
\( ( \tau \vec{b}-\kappa \vec{t} )\vec{a}=0\)
⟺\( \tau\vec{b}.\vec{a}-\kappa \vec{t}.\vec{a}=0\) (iii)
Here, (ii) shows that \( \vec{a}\) is perpendicular to \( \vec{n}\),
Hence, \( \vec{a}\) lies in rectifying plane.
Since, \( \vec{a}\) make \(\alpha \) with \( \vec{t}\), also \( \vec{a}\) make \( (90-\alpha) \) with \( \vec{b}\).
Therefore,
\( \vec{t}.\vec{a}= \cos \alpha \)and \( \vec{b}.\vec{a}= \cos( 90-\alpha )\)
Hence, from (iii), we have
\( \tau\vec{b}.\vec{a}-\kappa \vec{t}.\vec{a}=0\)
⟺\( \tau( \vec{b}.\vec{a} )-\kappa ( \vec{t}.\vec{a} )=0\)
⟺\( \tau \sin \alpha -\kappa \cos \alpha =0\)
⟺\( \frac{\kappa }{\tau }=\tan \alpha \), which is a constant.
This completes the proof.
The necessary and sufficient condition for a curve to be a helix is principal normal is always parallel to a fixed plane.
Solution
Let C be a space curve.
Then necessary and sufficient condition for C to be a helix is that it make a constant angle with a fixed direction.
In this case,
If \( \vec{a}\) is unit vector along direction of generator of C and \( \vec{t}\) is unit vector along tangent to C, then necessary and sufficient condition for C to be a helix is that
\( \vec{t}.\vec{a}=cos\alpha \) (i)
ing (i) w. r. to.\( s\), we get
\( \kappa \vec{n}.\vec{a}=0\)
\( \vec{n}.\vec{a}=0\) [In general \( \kappa \ne 0\)] (ii)
Here, (ii) shows that, principal normal \( \vec{n}\) is always perpendicular to a fixed vector , hence principal normal are always parallel to a fixed plane.
This completes the solution.
Necessary and sufficient condition for a curve to be a helix is \( [\vec{t} ',\vec{t} '',\vec{t} '''] =0 \).
Solution
Let \( C: \vec{r}=\vec{r}(s) \) is a space curve, then
\( [\vec{t} ',\vec{t} '',\vec{t} '''] = \kappa^5 \frac{d}{ ds} (\frac{ \tau}{ \kappa}) \)
Then necessary and sufficient condition for the curve C to be a helix is
\( \frac{ \kappa}{ \tau} =\text{constant} \)
or \( \frac{ \tau}{ \kappa} =\text{constant} \)
or \( \frac{d}{ ds} (\frac{ \tau}{ \kappa}) =0 \)
Hence, necessary and sufficient condition for a curve to be a helix is
\( [\vec{t} ',\vec{t} '',\vec{t} ''']=0 \)
This completes the solution.
Necessary and sufficient condition for a curve to be a helix is \( [\vec{b} ',\vec{b} '',\vec{b} '''] =0 \).
Solution
Let \( C: \vec{r}=\vec{r}(s) \) is a space curve, then
\( [\vec{b} ',\vec{b} '',\vec{b} '''] = \tau^5 \frac{d}{ ds} (\frac{ \kappa}{ \tau}) \)
Then necessary and sufficient condition for the curve C to be a helix is
\( \frac{ \kappa}{ \tau} =\text{constant} \)
or \( \frac{d}{ ds} (\frac{ \kappa}{ \tau}) =0 \)
Hence, necessary and sufficient condition for a curve to be a helix is
\( [\vec{b} ',\vec{b} '',\vec{b} ''']=0 \)
This completes the solution.
Curvature \( \kappa \) and torsion \( \tau\) of a helix \( C\) are in a constant ratio to the curvature \(\kappa_1\) of a plane curve \( C_1\) obtained by projecting \( C\) on a plane perpendicular to the axis of helix and are connected by relations
\( \kappa =\kappa_1 \sin ^2 \alpha \) and \( \tau=\kappa_1 \sin \alpha \cos\alpha \).
Proof
Let \( C:\vec{r}=\vec{r}( s )\) be a helix whose generators are taken along \( z\)-axis and projection of \( C\) on \(xy-\) plane is a plane curve \( C_1\).
Let arc length \( s\) of \( C\) is measured from suitable base point A such that
ArcAP = s,
Here, P (x,y,z) is a point on C and \( P_1( x,y,0 )\) is the corresponding point on \( C_1\)
Since, tangent to the helix makes a constant angle \(\alpha \) with direction \( \vec{a}\) of generator \( PP_1\), we have
\( \angle APP1 =\alpha \)
Now, from sector APP1
\( z=s \cos \alpha \) and \( s_1=s \sin \alpha \)
So
\( \frac{ds_1}{ds}=\sin \alpha \) (i)
Here, position of \( P\) on\( C\) is given by
\( \vec{r}=( x,y,s \cos \alpha )\)
Differentiating w. r. to.\( s\), we get
\( \vec{t}=( x',y', \cos \alpha )\)
Again, differentiating w. r. to \( s\), we get
\( \kappa \vec{n}=( x'',y'',0 )\)
Taking magnitude we get
\( \kappa =\sqrt{(x'')^2+( y'')^2}\) (ii)
Similarly, position of \( P_1 \) on \( C_1\) is
\(\vec{r}_1=( x,y,0 )\)
Differentiating w. r. to \( s\), we get
\(\vec{t}_1 \frac{ds_1}{ds}=( x',y',0 )\)
or\(\vec{t}_1 \sin \alpha =( x',y',0 )\)
Again, differentiating w. r. to \( s\), we get
\(\kappa_1 \vec{n}_1 \frac{ds_1}{ds} \sin \alpha =( x'',y'',0 )\)
or\(\kappa_1 \vec{n}_1 \sin ^2 \alpha =( x'',y'',0 )\)
Taking magnitude we get
\({{\kappa }_1}si{{n}^2}\alpha =\sqrt{(x'')^2+( y'')^2}\) (iii)
Dividing (ii) by (iii) we get
\( \kappa ={\kappa_1} \sin ^2 \alpha \) 1st relation established
Again for a helix, we have
\(tan\alpha =\frac{\kappa }{\tau }\)
or \( \tau={{\kappa }_1}sin\alpha .cos\alpha \) 2nd relation established
This completes the solution.
A helix which lies on a surface of circular cylinder is called a circular helix. It is also called right circular helix. The curvature and torsion of such helix are both constant.
Expression of Circular Helix
Let \( C:\vec{r}=\vec{r}( s )\) be a helix whose generators are taken alond \( z\)-axis and projection of \( C\) on xy plane describe a plane curve \( C_1\).We measure arc length \( s\) of \( C\) from a suitable base point \( A\) such that
ArcAP = s, arcAP1 = \(s_1\)
where
\( P( x,y,z )\) is on C and \( P_1( x,y,0 )\) is on \( C_1\).
Since, tangent to helix makes a constant angle \( \alpha \) with generator \( PP_1\),
\( \angle APP1 =\alpha \)
Now, from sector \( APP_1\)
\( z=s \cos \alpha \) and \( s_1=s \sin \alpha \)
So
\( s=\frac{s_1}{\sin \alpha }\)
Since \( C:\vec{r}=\vec{r}( s )\) is circular helix, \( C_1:x^2+y^2=a^2\)
And, parametric expression of \( C_1\) is
\( x=a \cos \theta ,y=a \sin \theta \)
Where \( \theta \)is angle that projection arc \( AP_1\) makes at origin Now,
\( s_1=a\theta \) from sector \( OAP_1\)
Thus,
\( z=s \cos \alpha =\frac{s_1}{sin\alpha } \cos \alpha =a\theta \cot \alpha \)
Hence, expression of circular helix is
\( \vec{r}=( x,y,z) \)
or \( \vec{r}=( a \cos \theta ,a \sin \theta ,a \theta \cot \alpha) \)
This completes the solution.
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